Question

F(x)=8th root of (8-x^2)sec(4x)... cant figure out the derivitive of this.. any help?

Answer 1

try splitting it up into two parts: [(8-x^2)^(1/8)][(sec(4x))^(1/8)] Sorry if all those parens get confusing, but basically its the two seperate parts being raised to the 1/8 power and are multiplied.

Answer 2

sorry the sec(4x) isnt in the root. its being mulitplied by the 8th root of (8-x^2)

Answer 3

Well either way, its still just f(x) = ab, so \[\prime\](x) = (a)(db) + (b)(da)

Answer 4

\[f \prime(x)*\]

Answer 5

what would the der. of the 8th root of (8-x^2) be? i got how the formulas to use are, but i dont know what im doing wrong here...

Answer 6

Thats the chain rule. So it would be (1/8)*[(8-x^2)^(-7/8)]*(-2x)

Answer 7

ok i see where i went wrong.. thanks a lot